Question

A block of mass m moving at speed `upsilon` collides wilth another block of mass 2 m at rest. The lighter block comes to rest after the collision. Find the coefficient of restitution.
A. `0.5`
B. `0.4`
C. `0.6`
D. `0.8`

Answer 1

Correct Answer - A
Let `upsilon_(1)` be the velocity of mass `2m` after collision. According to law of conservation of linear momentum .
`m upsilon+2mxx0=mxx0+2m upsilon_(1)` or `upsilon_(1)=upsilon//2`
Coefficient of restitution `(e)`
`=("velocity of separation")/("velocity of approach")=(upsilon_(1)-0)/(upsilon-0)=(upsilon_(1))/(upsilon)`
`(upsilon//2)/(upsilon)=0.5`